SOLVING WORD PROBLEMS INVOLVING DISTANCE SPEED AND TIME

Problem 1 :

The distance between two stations A and B is 192 km. Traveling by a fast train takes 48 minutes less than another train. Calculate the speed of the fast train if the speeds of two trains differ by 20 km/hr

Solution :

Distance between two stations A and B  =  192 km

Fast train takes 48 minutes less then the time taken by the slow train.

Let x be the speed taken by the fast train.

Speeds of two trains differ by 20 km/hr

So speed of slow train is x – 20.

Time  =  Distance/speed

Let t₁ and t₂ be the time taken by the faster train and slower train respectively.

Let  be the time taken by the slow train

t₁  =  192 / x

t₂  =  192 / (x – 20)

48 / 60 ==>  4/5 hours

t_{2 -} t₁  =  4/5

[192 / (x -20) – 192 / x] = 4/5

192[ (x – x + 20)/x (x - 20) ]  =  4/5

19200  =  4x² – 80x

Dividing the entire equation by 4, we get

4800  =  x² – 20x

x² – 20x – 4800  =  0

(x – 60) (x + 40)  =  0

So, speed of the faster train = 60 km/hr

Speed of slow train  =  x – 20

  =  60 - 20

  =  40 km/hr

Speed of the faster train  =  60 km/hr.

Speed of the slow train  =  40 km/hr.

Problem 2 :

A train covers a distance of 300 km at a certain average speed. If its speed was decreased by 10 km/hr, the journey would take 1 hour longer. What is the average speed.

Solution :

Let x be the average speed of the train

If its speed was decreased by 10 km/hr, the journey would take 1 hour longer.

So, x – 10 be the decreased speed

Time  =  Distance / Speed

Let t₁ and t₂ time taken to cover the distance in the speed of x and x - 10 km/hr.

t₁  =  300/x

t₂  =  300/(x – 10)

t₁ – t₂  =  1

( 300/x ) -  ( 300/(x - 10) )  =  1

300 [1/(x - 10) – 1/x] = 1

300 [x –  x + 10]/[x(x -10)]  =  1

3000/(x² – 10x) = 1

3000  =  x² – 10x

x² – 10x  =  3000

x² – 10x – 3000  =  0

(x + 50)(x – 60)  =  0

So speed of the 60 km/hr.

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